$\lim_{x\to\infty}\dfrac{5x^2-8}{7x-4x^2}=?$ Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{5}{4}$ (Choice B) B $0$ (Choice C) C $\dfrac{5}{7}$ (Choice D) D $-\infty$
$\lim_{x\to\infty} 5x^2-8=\infty$ and $\lim_{x\to\infty} 7x-4x^2=-\infty$, so $\lim_{x\to\infty}\dfrac{5x^2-8}{7x-4x^2}$ results in the indeterminate form $\dfrac{\infty}{-\infty}$. We should use l'Hôpital's rule. $\begin{aligned} &\phantom{=}\lim_{x\to\infty}\dfrac{5x^2-8}{7x-4x^2} \\\\ &=\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[5x^2-8\right]}{\dfrac{d}{dx}[7x-4x^2]} \gray{\text{l'Hôpital's rule}} \\\\ &=\lim_{x\to\infty}\dfrac{10x}{7-8x} \\\\ &=\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[10x\right]}{\dfrac{d}{dx}[7-8x]} \gray{\text{l'Hôpital's rule}} \\\\ &=\lim_{x\to\infty}\dfrac{10}{-8} \\\\ &=-\dfrac{5}{4} \end{aligned}$ Note that we used l'Hôpital's rule twice, because the first time we used it, we ended with the indeterminate form $\dfrac{\infty}{-\infty}$ too. Also note that we were only able to use l'Hôpital's rule because the limit $\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[10x\right]}{\dfrac{d}{dx}[7-8x]}$ can actually be determined. In conclusion, $\lim_{x\to\infty}\dfrac{5x^2-8}{7x-4x^2}=-\dfrac{5}{4}$.